Relativistic Electrodynamics

Continuity Equation

Recall the continuity equation $$\nabla\cdot\vec{J}=-\frac{\partial \rho}{\partial t}.$$ In terms of the four-current density, the left hand side is $$\nabla\cdot\vec{J}=\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}+\frac{\partial J_z}{\partial z}=\sum^3_{i=1}\frac{\partial J^i}{\partial x^i},$$ while the right hand side is $$\frac{\partial \rho}{\partial t}=\frac{1}{c}\frac{\partial J^0}{\partial t}=\frac{\partial J^0}{\partial x^0}.$$ Thus, the continuity equation in terms of the four-current density is $$\bbox[5px, border: 2px solid #666]{ \frac{\partial J^\mu}{\partial x^\mu} = 0. }$$ As this is also the four-dimensional divergence of \(J^\mu\), the continuity equation implies the four-current density is divergenceless.

Maxwell's Equations in Terms of Field Tensor

Recall the Maxwell's equations \begin{align} \nabla\cdot\vec{E} &= \frac{\rho}{\epsilon}\tag{1}\\ \nabla\times\vec{B} &= \mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}\tag{2}\\ \nabla\cdot{B} &= 0\tag{3}\\ \nabla\times\vec{E} &= -\frac{\partial \vec{B}}{\partial t}\tag{4}. \end{align} For equation (1), let us examine \(\frac{\partial F^{0\nu}}{\partial x^\nu}\) \begin{align} \frac{\partial F^{0\nu}}{\partial x^\nu}&=\frac{\partial F^{00}}{\partial x^0} +\frac{\partial F^{01}}{\partial x^1} +\frac{\partial F^{02}}{\partial x^2} +\frac{\partial F^{03}}{\partial x^3}\\ &= \frac{1}{c}\left(\frac{\partial E_x}{\partial x^0}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}\right)\\ &= \frac{1}{c}(\nabla\cdot\vec{E})\\ &= \mu_0 c\rho\\ &= \mu_0J^0. \end{align} Thus, $$\frac{\partial F^{0\nu}}{\partial x^\nu} = \mu_0J^0$$ is equivalent to equation (1).

For equation (2), let us examine \(\frac{\partial F^{1\nu}}{\partial x^\nu}\) \begin{align} \frac{\partial F^{1\nu}}{\partial x^\nu} &= \frac{\partial F^{10}}{\partial x^0} + \frac{\partial F^{11}}{\partial x^1} + \frac{\partial F^{12}}{\partial x^2} + \frac{\partial F^{13}}{\partial x^3}\\ &= -\frac{1}{c^2}\frac{\partial E_x}{\partial t} + \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z}\\ &= \left(-\frac{1}{c^2}\frac{\partial E}{\partial t}+\nabla\times\vec{B}\right)_x\\ &= \mu_0J_x \\ &= \mu_0J^1. \end{align} Similarly for the \(\frac{\partial F^{2\nu}}{\partial x^\nu}\) and \(\frac{\partial F^{3\nu}}{\partial x^\nu}\). Thus, $$\frac{\partial F^{i\nu}}{\partial x^\nu} = \mu_0J^i$$ is equivalent to the equation (2).

For equation (3), let us examine \(\frac{\partial G^{0\nu}}{\partial x^\nu}\) \begin{align} \frac{\partial G^{0\nu}}{\partial x^\nu} &= \frac{\partial G^{00}}{\partial x^0} +\frac{\partial G^{01}}{\partial x^1} +\frac{\partial G^{02}}{\partial x^2} +\frac{\partial G^{03}}{\partial x^3}\\ &= \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} \\ &= \nabla\cdot\vec{B} = 0. \end{align} Thus, $$\frac{\partial G^{0\nu}}{\partial x^\nu}=0$$ is equivalent to equation (3).

For equation (4), let us examine \(\frac{\partial G^{1\nu}}{\partial x^\nu}\) \begin{align} \frac{\partial G^{1\nu}}{\partial x^\nu} &= \frac{\partial G^{10}}{\partial x^0} + \frac{\partial G^{11}}{\partial x^1} + \frac{\partial G^{12}}{\partial x^2} + \frac{\partial G^{13}}{\partial x^3} \\ &= -\frac{1}{c}\frac{\partial B_x}{\partial t}-\frac{1}{c}\frac{\partial E_z}{\partial y}+\frac{1}{c}\frac{\partial E_y}{\partial z}\\ &= -\frac{1}{c}\left(\frac{\partial \vec{B}}{\partial t}+\nabla\times\vec{E}\right)_x\\ &= 0. \end{align} Similarly for the \(\frac{\partial G^{2\nu}}{\partial x^\nu}\) and \(\frac{\partial G^{3\nu}}{\partial x^\nu}\). Thus, $$\frac{\partial G^{i\nu}}{\partial x^\nu}=0$$ is equivalent to equation(4).

Therefore, the Maxwell's equations in terms of the four-current density are \begin{cases} \frac{\partial F^{\mu\nu}}{\partial x^\nu}=\mu_0J^\mu\\ \frac{\partial G^{\mu\nu}}{\partial x^\nu}=0 \end{cases}

Maxwell's Equations in terms of 4-vector Potential

Substitute $$F^{\mu\nu}=\frac{\partial A^\nu}{\partial x_\mu}-\frac{\partial A^\mu}{\partial x_\nu}$$ into $$\frac{\partial F^{\mu\nu}}{\partial x^\nu}=\mu_0J^\mu,$$ we have $$\frac{\partial}{\partial x_\mu}\left(\frac{\partial A^\nu}{\partial x^\nu}\right)-\frac{\partial}{\partial x_\nu}\left(\frac{\partial A^\mu}{\partial x^\nu}\right)=\mu_0J^\mu.\tag{5}$$ As \(F^{\mu\nu}\) is gauge invariant (as \(\vec{E}\) and \(\vec{B}\) are gauge invariant), the gauge transformation $$A^\mu\to A^{\mu}'=A^\mu + \frac{\partial \lambda}{\partial x_\mu}$$ leaves \(F^{\mu\nu}\) unchanged. The Lorentz gauge condiiton $$\nabla\cdot\vec{A}=-\frac{1}{c^2}\frac{\partial V}{\partial t}$$ in terms of 4-vector potential is $$\frac{\partial A^\nu}{\partial x^\nu}=0.$$ Therefore, (5) reduces to $$\bbox[5px, border: 2px solid #666]{ \Box^2A^\mu = -\mu_0 J^\mu, }$$ where \(\Box\) is the d'Alembertian defined as $$\Box^2:=\frac{\partial}{\partial x_\nu}\frac{\partial}{\partial x^\nu}=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}p$$